auto increment inside group

问题

I have a dataframe:

df = pd.DataFrame.from_dict({
    'product': ('a', 'a', 'a', 'a', 'c', 'b', 'b', 'b'),
    'sales': ('-', '-', 'hot_price', 'hot_price', '-', 'min_price', 'min_price', 'min_price'),
    'price': (100, 100, 50, 50, 90, 70, 70, 70),
    'dt': ('2020-01-01 00:00:00', '2020-01-01 00:05:00', '2020-01-01 00:07:00', '2020-01-01 00:10:00', '2020-01-01 00:13:00', '2020-01-01 00:15:00', '2020-01-01 00:19:00', '2020-01-01 00:21:00')
})

  product      sales  price                   dt
0       a          -    100  2020-01-01 00:00:00
1       a          -    100  2020-01-01 00:05:00
2       a  hot_price     50  2020-01-01 00:07:00
3       a  hot_price     50  2020-01-01 00:10:00
4       c          -     90  2020-01-01 00:13:00
5       b  min_price     70  2020-01-01 00:15:00
6       b  min_price     70  2020-01-01 00:19:00
7       b  min_price     70  2020-01-01 00:21:00

I need the next output:

  product      sales  price                   dt  unique_group
0       a          -    100  2020-01-01 00:00:00             0
1       a          -    100  2020-01-01 00:05:00             0
2       a  hot_price     50  2020-01-01 00:07:00             1
3       a  hot_price     50  2020-01-01 00:10:00             1
4       c          -     90  2020-01-01 00:13:00             2
5       b  min_price     70  2020-01-01 00:15:00             3
6       b  min_price     70  2020-01-01 00:19:00             3
7       b  min_price     70  2020-01-01 00:21:00             3

How I do it:

unique_group = 0
df['unique_group'] = unique_group
for i in range(1, len(df)):
    current, prev = df.loc[i], df.loc[i - 1]
    if not all([
        current['product'] == prev['product'],
        current['sales'] == prev['sales'],
        current['price'] == prev['price'],
    ]):
        unique_group += 1
    df.loc[i, 'unique_group'] = unique_group

Is it possible to do it without iteration? I tried using cumsum(), shift(), ngroup(), drop_duplicates() but unsuccessfully.

回答1:

IIUC, GroupBy.ngroup:

df['unique_group'] = df.groupby(['product', 'sales', 'price'],sort=False).ngroup()
print(df)

  product      sales  price                   dt  unique_group
0       a          -    100  2020-01-01 00:00:00             0
1       a          -    100  2020-01-01 00:05:00             0
2       a  hot_price     50  2020-01-01 00:07:00             1
3       a  hot_price     50  2020-01-01 00:10:00             1
4       c          -     90  2020-01-01 00:13:00             2
5       b  min_price     70  2020-01-01 00:15:00             3
6       b  min_price     70  2020-01-01 00:19:00             3
7       b  min_price     70  2020-01-01 00:21:00             3

this works either way, even if the data frame is not ordered

Another approach

this works with the ordered data frame

cols = ['product','sales','price']
df['unique_group'] = df[cols].ne(df[cols].shift()).any(axis=1).cumsum().sub(1)

回答2:

Another option which might be a bit faster than groupby:

df['unique_group'] = (~df.duplicated(['product','sales','price'])).cumsum() - 1

Output:

  product      sales  price                   dt  unique_group
0       a          -    100  2020-01-01 00:00:00             0
1       a          -    100  2020-01-01 00:05:00             0
2       a  hot_price     50  2020-01-01 00:07:00             1
3       a  hot_price     50  2020-01-01 00:10:00             1
4       c          -     90  2020-01-01 00:13:00             2
5       b  min_price     70  2020-01-01 00:15:00             3
6       b  min_price     70  2020-01-01 00:19:00             3
7       b  min_price     70  2020-01-01 00:21:00             3

来源：https://stackoverflow.com/questions/60727830/auto-increment-inside-group

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pandas

grouping